Question

What is the limit of $\sin \left( \dfrac{1}{x} \right)$ as $x$ approaches $0$ ?

Answer 1

For these kinds of questions, we should know the concept of limits quite thoroughly. We should know a lot of theorems which are involved. In this particular question, we will solve It using the sandwich theorem. The basic intuition of the Sandwich theorem is that a particular function is sandwiched or completely squeezed between two other different functions. So the value of limits of other functions at some particular value of $x$ would be the value of the limit of the sandwiched function at that value of $x$.

Complete step by step solution:
Let $f\left( x \right),g\left( x \right),h\left( x \right)$ be real function such that $f\left( x \right)\le g\left( x \right)\le h\left( x \right),\forall \text{ }x$ in the common domain.
If $\displaystyle \lim_{x \to a}f\left( x \right)=l=\displaystyle \lim_{x \to a}h\left( x \right)$ then $\displaystyle \lim_{x \to a}g\left( x \right)=l$.
This is what the Sandwich theorem states. Here, the function which is sandwiched between two functions is $g\left( x \right)$. And the functions which are sandwiching $g\left( x \right)$ are $h\left( x \right),f\left( x \right)$.
Now let us use this to solve our question.
We know that the range of the sin trigonometric function is $\left[ -1,1 \right]$ . It means no matter the value of a function , the sin of it will always lie between $\left[ -1,1 \right]$ .
Let us apply this.
Upon applying , we get the following :
$\Rightarrow -1\le \sin \left( \dfrac{1}{x} \right)\le 1$
So here, we are sandwiching our function $\sin \left( \dfrac{1}{x} \right)$ between $y=-1$ and $y=1$ .
Now let us apply the limit.
Upon doing so, we get the following :
$\begin{aligned} & \Rightarrow -1\le \sin \left( \dfrac{1}{x} \right)\le 1 \\\ & \Rightarrow \displaystyle \lim_{x \to 0}-1\le \displaystyle \lim_{x \to 0}\sin \left( \dfrac{1}{x} \right)\le \displaystyle \lim_{x \to 0}1 \\\ & \Rightarrow -1\le \displaystyle \lim_{x \to 0}\sin \left( \dfrac{1}{x} \right)\le 1 \\\ \end{aligned}$
Since $y=-1,y=1$ are not functions of $x$ , upon applying the limit, nothing will change. We have to find out $\displaystyle \lim_{x \to 0}\sin \left( \dfrac{1}{x} \right)$.
We can clearly see that the value of the limit of $\sin \left( \dfrac{1}{x} \right)$ as $x$ approaches $0$ lies between $\left[ -1,1 \right]$ . It is not merely lying between them, it is oscillating between them . As it approaches $0$ , it oscillates infinitely fast.
Graph :

Note: We should learn all the theorems involved in calculus not just limits. Calculus is a vast topic and there is every possibility for a question which has many concepts in it. We should be exposed to different level questions so as to get the idea for the question which is asked in the exam. In limits, we should know the different ways to find out the limit of functions. Limit by expansion, limit using theorems, L-Hospital Rule.